1
$\begingroup$

This question has to be divided into the following parts:

  1. The definition of Day convolution in nlab

    To define Day convolution, it assumes that $V$ be a closed symmetric monoidal category with all small limits and colimits, and $C$ be a monoidal category.

    see https://ncatlab.org/nlab/show/Day+convolution#definition

    Notice that nlab doesn't says that $C$ must be symmetric.

  2. Day convolution form a monoidal category in nlab

    see https://ncatlab.org/nlab/show/Day+convolution#DayConvolutionYieldsMonoidalCategoryStructure

    That means, if we there is a tensor unit $y(I)$, then the category $([C,V], ⊗_{Day}, y(I))$ form a monoidal category automatically.

    Notice that nlab doesn't say that $C$ must be symmetric.

  3. The definition of Day convolution in wikipedia

    To define Day convolution, it assumes that $C$ be a symmetric monoidal category. (Of course, $V$ must be monoidal category, because enriched)

    see https://en.wikipedia.org/wiki/Day_convolution

    Notice that wikipedia doesn't say that $V$ must be symmetric.

  4. Day convolution form a monoidal category in wikipedia

    It says that

    If the category $V$ is a symmetric monoidal closed category, we can show this defines an associative monoidal product.

    see https://en.wikipedia.org/wiki/Day_convolution

    Since a monoidal category must satisfy associative law, that means if we expect that the category $([C,V], ⊗_{Day}, y(I))$ form a monoidal category, then $V$ must be symmetric, i.e. $C$ and $V$ are both symmetric monoidal category.

    It also provides a proof for this associative law, in which, it seems that the two symmetric /commutative laws be used.

My questions are:

  1. Why the definition of Day convolution in nlab and wikipedia are different?

    I mean that, to define Day convolution, why nlab require $V$ to be a symmetric monoidal category, but wikipedia doesn't require symmetric on $V$ and vice versa...

  2. Why the condition of "Day convolution form a monoidal category" in nlab and wikipedia are different?

    I mean that, to form a monoidal category under Day convolution, why wikipedia require both $C$ and $V$ are symmetric, but nlab doesn't require this condition?

  3. Why Day convolution need some sort of "symmetric" property?

    I didn't see any symmetry intuition from this Day convolution formula:

    $F*G = \int^{x,y \in C} C(x \otimes y, -) \otimes Fx \otimes Gy$

PS: I apologize if the question is silly, I'm a category theory beginner, but this definition make me confusion...

Very thanks.

$\endgroup$
2
$\begingroup$

The description on the nLab is correct: $\mathscr C$ does not need to be symmetric, but $\mathscr V$ does. If $\mathscr C$ is symmetric, then the Day convolution tensor product on $[\mathscr C, \mathscr V]$ will also be symmetric. Wikipedia actually does require $\mathscr V$ to be symmetric, but delays stating this to establish why symmetry is important: it's necessary for the induced tensor product to be associative (and hence be monoidal). This matches Day's original setting.

As of the time of writing, Wikipedia does state that $\mathscr C$ should be symmetric, but this is unnecessary. Anyone can edit Wikipedia, so this could easily be addressed.

$\endgroup$
4
  • $\begingroup$ Thanks. I found the wikipedia has been modified. Now to define Day convolution, we just require $(C, \otimes_c)$ be a monoidal category and $(V, \otimes)$ be a symmetric monoidal closed category, right? But the proof of monoidal associative law in wikipedia still need $\otimes_c$ symmetric? see 5th isomorphism, it seems need $\otimes_c$ symmetric? But in nlab, this associative law should not require $\otimes_c$ symmetric. $\endgroup$
    – chansey
    Jun 21 '20 at 15:25
  • $\begingroup$ You're right: the current proof on Wikipedia is a little odd. If you take a look at the proof in this answer, or in the proof of Proposition 6.2.1 in Coend calculus, you'll see that symmetry is not required. (It's a helpful exercise to prove this yourself if you haven't already tried, though.) $\endgroup$
    – varkor
    Jun 21 '20 at 15:30
  • $\begingroup$ I have read the proof in this answer, it's great, thanks! So the proof of the associative law doesn't need $\otimes_C$ be symmetric, it just need ninja Yoneda lemma, associativity of $\otimes_C$ and symmetric of $\otimes_V$. So the answer of 3rd question shoube be that the symmetric of $\otimes_V$ can make a monoidal category. $\endgroup$
    – chansey
    Jun 21 '20 at 16:24
  • 1
    $\begingroup$ Yes, that's right :) $\endgroup$
    – varkor
    Jun 21 '20 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.